$\displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$\displaystyle \ \ \ \sin \theta =y$

$\displaystyle \ \ \ \cos \theta =x$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,$

$\displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$

$\displaystyle \begin{array}{l}\ \ \ \text{But it is clear that, and }{y}'\ \text{have }y\ \text{opposite signs}\\\ \ \text{ }{x}'\ \text{and }x\ \text{have the same sign,}\end{array}$

$\displaystyle \therefore {y}'=-y\ \text{and }{x}'=x.$

$\displaystyle \therefore \sin (-\theta )={y}'=-y=-\sin \theta$

$\displaystyle \ \ \ \cos (-\theta )={x}'=x=\cos \theta$

$\displaystyle \ \ \ \tan (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{y}{x}=-\tan \theta$

$\displaystyle \ \ \ \cot (-\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=-\cot \theta$

$\displaystyle \ \ \ \sec (-\theta )=\frac{1}{{{x}'}}=\frac{1}{x}=\sec \theta$

$\displaystyle \ \ \ \operatorname{cosec}(-\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta$

Slider ကို ေ႐ႊ႕ၾကည့္ပါ။

$\displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$\displaystyle \ \ \ \sin \theta =y$

$\displaystyle \ \ \ \cos \theta =x$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$

$\displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$

$\displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the fourth quadrant}\text{.}$

$\displaystyle \therefore {y}'=-x\ \text{and }{x}'=y.$

$\displaystyle \ \ \ \sin (270{}^\circ +\theta )={y}'=-x=-\cos \theta$

$\displaystyle \ \ \ \cos (270{}^\circ +\theta )={x}'=y=\sin \theta$

$\displaystyle \ \ \ \tan (270{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{x}}{{y}}=\cot \theta$

$\displaystyle \ \ \ \cot (270{}^\circ +\theta )=\frac{{{x}'}}{{{y}'}}=-\frac{{y}}{{x}}=\tan \theta$

$\displaystyle \ \ \ \sec (270{}^\circ +\theta )=\frac{1}{{{x}'}}=\frac{1}{y}=\operatorname{cosec}\theta$

$\displaystyle \ \ \ \operatorname{cosec}(270{}^\circ +\theta )=\frac{1}{{{y}'}}=-\frac{1}{x}=-\sec\theta$

θ တန္ဖိုး ရိုက္ထည့္ပါ။

$\displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$\displaystyle \ \ \ \sin \theta =y$

$\displaystyle \ \ \ \cos \theta =x$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$

$\displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$

$\displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the third quadrant}\text{.}$

$\displaystyle \therefore {y}'=-x\ \text{and }{x}'=-y.$

$\displaystyle \ \ \ \sin (270{}^\circ -\theta )={y}'=-x=-\cos \theta$

$\displaystyle \ \ \ \cos (270{}^\circ -\theta )={x}'=-y=-\sin \theta$

$\displaystyle \ \ \ \tan (270{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=\frac{{-x}}{{-y}}=\cot \theta$

$\displaystyle \ \ \ \cot (270{}^\circ -\theta )=\frac{{{x}'}}{{{y}'}}=\frac{{-y}}{{-x}}=\tan \theta$

$\displaystyle \ \ \ \sec (270{}^\circ -\theta )=\frac{1}{{{x}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta$

$\displaystyle \ \ \ \operatorname{cosec}(270{}^\circ -\theta )=\frac{1}{{{y}'}}=-\frac{1}{x}=-\sec\theta$

$\displaystyle \theta$ တန္ဖိုး ရိုက္ထည့္ပါ။

$\displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$\displaystyle \ \ \ \sin \theta =y$

$\displaystyle \ \ \ \cos \theta =x$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \ \ \ \sin (90{}^\circ -\theta )=x=\cos \theta$

$\displaystyle \ \ \ \cos (90{}^\circ -\theta )=y=\sin \theta$

$\displaystyle \ \ \ \tan (90{}^\circ -\theta )=\frac{x}{y}=\cot \theta$

$\displaystyle \ \ \ \cot (90{}^\circ -\theta )=\frac{y}{x}=\tan \theta$

$\displaystyle \ \ \ \sec (90{}^\circ -\theta )=\frac{1}{y}=\operatorname{cosec}\theta$

$\displaystyle \ \ \ \operatorname{cosec}(90{}^\circ -\theta )=\frac{1}{x}=\sec\theta$

$\displaystyle \theta$ တန္ဖိုး ရိုက္ထည့္ပါ။

$\displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$\displaystyle \ \ \ \sin \theta =y$

$\displaystyle \ \ \ \cos \theta =x$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,$

$\displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$

$\displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the fourth quadrant}\text{.}$

$\displaystyle \therefore {y}'=-y\ \text{and }{x}'=x.$

$\displaystyle \therefore \sin (360{}^\circ -\theta )={y}'=-y=-\sin \theta$

$\displaystyle \ \ \ \cos (360{}^\circ -\theta )={x}'=x=\cos \theta$

$\displaystyle \ \ \ \tan (360{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{y}}{{x}}=-\tan \theta$

$\displaystyle \ \ \ \cot (360{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{{x}}{{y}}=-\cot \theta$

$\displaystyle \ \ \ \sec (360{}^\circ -\theta )=\frac{1}{{{x}'}}=\frac{1}{x}=\sec \theta$

$\displaystyle \ \ \ \operatorname{cosec}(360{}^\circ -\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta$

Dynamic Presentation

$\displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$\displaystyle \ \ \ \sin \theta =y$

$\displaystyle \ \ \ \cos \theta =x$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,$

$\displaystyle \ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}$

$\displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the third quadrant}\text{.}$

$\displaystyle \therefore {y}'=-y\ \text{and }{x}'=-x.$

$\displaystyle \therefore \sin (180{}^\circ +\theta )={y}'=-y=-\sin \theta$

$\displaystyle \ \ \ \cos (180{}^\circ +\theta )={x}'=-x=-\cos \theta$

$\displaystyle \ \ \ \tan (180{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=\frac{{-y}}{{-x}}=\tan \theta$

$\displaystyle \ \ \ \cot (180{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=\frac{{-x}}{{-y}}=\cot \theta$

$\displaystyle \ \ \ \sec (180{}^\circ +\theta )=\frac{1}{{{x}'}}=-\frac{1}{x}=-\sec \theta$

$\displaystyle \ \ \ \operatorname{cosec}(180{}^\circ +\theta )=\frac{1}{{{y}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta$

Dynamic Presentation

$\displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$\displaystyle \ \ \ \sin \theta =y$

$\displaystyle \ \ \ \cos \theta =x$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$

$\displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$

$\displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the second quadrant}\text{.}$

$\displaystyle \therefore {y}'=x\ \text{and }{x}'=-y.$

$\displaystyle \ \ \ \sin (90{}^\circ +\theta )={y}'=x=\cos \theta$

$\displaystyle \ \ \ \cos (90{}^\circ +\theta )={x}'=-y=-\sin \theta$

$\displaystyle \ \ \ \tan (90{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=\cot \theta$

$\displaystyle \ \ \ \cot (90{}^\circ +\theta )=\frac{{{x}'}}{{{y}'}}=-\frac{y}{x}=\tan \theta$

$\displaystyle \ \ \ \sec (90{}^\circ +\theta )=\frac{1}{{{x}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta$

$\displaystyle \ \ \ \operatorname{cosec}(90{}^\circ +\theta )=\frac{1}{{{y}'}}=\frac{1}{x}=sec\theta$

$\displaystyle \theta$ တန္ဖိုး႐ိုက္ထည့္ၾကည့္ပါ။

$\displaystyle \begin{array}{l}\ \ \ \text{In}\ \vartriangle PON,\\\\\ \ \ \sin \theta =y\\\\\ \ \ \cos \theta =x\end{array}$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \begin{array}{l}\ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,\\\\\ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}\\\\\ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the second quadrant}\text{.}\\\\\therefore {y}'=y\ \text{and }{x}'=-x.\\\\\therefore \sin (180{}^\circ -\theta )={y}'=y=\sin \theta \\\\\ \ \ \cos (180{}^\circ -\theta )={x}'=-x=-\cos \theta \end{array}$

$\displaystyle \ \ \ \tan (180{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{y}{x}=-\tan \theta$

$\displaystyle \ \ \ \cot (180{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=-\cot \theta$

$\displaystyle \ \ \ \sec (180{}^\circ -\theta )=\frac{1}{{{x}'}}=-\frac{1}{x}=-\sec \theta$

$\displaystyle \ \ \ \operatorname{cosec}(180{}^\circ -\theta )=\frac{1}{{{y}'}}=\frac{1}{y}=\operatorname{cosec}\theta$

Dynamic Presentation

(a) $\displaystyle \ \ \ \frac{{\tan \theta -1+\sec \theta }}{{\tan \theta +1-\sec \theta }}$

$\displaystyle \text{=}\frac{{\tan \theta +\sec \theta -1}}{{\tan \theta -\sec \theta +1}}$

$\displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)-\left( {{{{\sec }}^{2}}\theta -{{{\tan }}^{2}}\theta } \right)}}{{\tan \theta -\sec \theta +1}}\$
$\displaystyle \ \ \ \left[ {\because {{{\tan }}^{2}}\theta +1={{{\sec }}^{2}}\theta } \right]$

$\displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)-\left( {\sec \theta +\tan \theta } \right)\left( {\sec \theta -\tan \theta } \right)}}{{\tan \theta +1-\sec \theta }}$

$\displaystyle \text{=}\frac{{\left( {\tan \theta +\sec \theta } \right)\left( {1-\sec \theta +\tan \theta } \right)}}{{1-\sec \theta +\tan \theta }}$

$\displaystyle =\tan \theta +\sec \theta$

$\displaystyle =\frac{{\sin \theta }}{{\cos \theta }}+\frac{1}{{\cos \theta }}$

$\displaystyle =\frac{{1+\sin \theta }}{{\cos \theta }}$

(b) $\displaystyle \ \ \ \frac{{1-\cos x}}{{1+\cos x}}$

$\displaystyle =\frac{{1-\cos x}}{{1+\cos x}}\times \frac{{1-\cos x}}{{1-\cos x}}$

$\displaystyle =\frac{{{{{\left( {1-\cos x} \right)}}^{2}}}}{{1-{{{\cos }}^{2}}x}}$

$\displaystyle =\frac{{1-2\cos x+{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}$

$\displaystyle =\frac{1}{{{{{\sin }}^{2}}x}}-\frac{2}{{\sin x}}\cdot \frac{{\cos x}}{{\sin x}}+\frac{{{{{\cos }}^{2}}x}}{{{{{\sin }}^{2}}x}}$

$\displaystyle ={{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x+{{\cot }^{2}}x$

$\displaystyle ={{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x+{{\operatorname{cosec}}^{2}}x-1\$
$\displaystyle \ \ \ \left[ {\because 1+{{{\cot }}^{2}}x={{{\operatorname{cosec}}}^{2}}x} \right]$

$\displaystyle =2{{\operatorname{cosec}}^{2}}x-2\operatorname{cosec}x\cot x-1$

(c)$\displaystyle \ \ \ (1+\tan x-\sec x)(1+\cot x+\operatorname{cosec}x)$

$\displaystyle =\ \left( {1+\frac{{\sin x}}{{\cos x}}-\frac{1}{{\cos x}}} \right)\left( {1+\frac{{\cos x}}{{\sin x}}+\frac{1}{{\sin x}}} \right)$

$\displaystyle =\frac{{\cos x+\sin x-1}}{{\cos x}}\ \times \frac{{\sin x+\cos x+1}}{{\sin x}}$

$\displaystyle =\frac{{{{{(\cos x+\sin x)}}^{2}}-1}}{{\sin x\cos x}}$

$\displaystyle =\frac{{{{{\cos }}^{2}}x+2\sin x\cos x+{{{\sin }}^{2}}x-1}}{{\sin x\cos x}}$

$\displaystyle =\frac{{{{{\cos }}^{2}}x+{{{\sin }}^{2}}x+2\sin x\cos x-1}}{{\sin x\cos x}}$

$\displaystyle =\frac{{1+2\sin x\cos x-1}}{{\sin x\cos x}}$

$\displaystyle =2$

In $\displaystyle \vartriangle ABC$,

$\displaystyle \sin \theta =\frac{a}{c}\ \ \ \ \ \ \ \ \cos \theta =\frac{b}{c}\ \ \ \ \ \ \ \ \tan \theta =\frac{a}{b}$

$\displaystyle \cot \theta =\frac{b}{a}\ \ \ \ \ \ \ \ \sec \theta =\frac{c}{b}\ \ \ \ \ \ \ \ \operatorname{cosec}\theta =\frac{c}{a}$

$\displaystyle \tan \theta =\frac{a}{b}=\frac{{\displaystyle \frac{a}{c}}}{{\displaystyle \frac{b}{c}}}=\frac{{\sin \theta }}{{\cos \theta }}$

$\displaystyle \cot \theta =\frac{b}{a}=\frac{{\displaystyle \frac{b}{c}}}{{\displaystyle \frac{a}{c}}}=\frac{{\cos \theta }}{{\sin \theta }}$

$\displaystyle \cot \theta =\frac{b}{a}=\frac{{\displaystyle \frac{b}{b}}}{{\displaystyle \frac{a}{b}}}=\frac{1}{{\tan \theta }}$

$\displaystyle \sec \theta =\frac{c}{b}=\frac{{\displaystyle \frac{c}{c}}}{{\displaystyle \frac{b}{c}}}=\frac{1}{{\cos \theta }}$

$\displaystyle \operatorname{cosec}\theta =\frac{c}{a}=\frac{{\displaystyle \frac{c}{c}}}{{\displaystyle \frac{a}{c}}}=\frac{1}{{\sin \theta }}$

$\displaystyle \begin{array}{l}\ \ \ \text{By Pythagoras'}\ \text{ theorem,}\\\\\ \ \ {{a}^{2}}+{{b}^{2}}={{c}^{2}}\end{array}$

$\displaystyle \therefore \frac{{{{a}^{2}}}}{{{{c}^{2}}}}+\frac{{{{b}^{2}}}}{{{{c}^{2}}}}=1$

$\displaystyle \therefore {{\left( {\frac{a}{c}} \right)}^{2}}+{{\left( {\frac{b}{c}} \right)}^{2}}=1$

$\displaystyle \therefore \ \begin{array}{|l|} \hline \displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\ \hline\end{array}$

$\displaystyle \ \ \ \text{Since}\ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$

$\displaystyle \ \ \ \frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}=\frac{1}{{{{{\cos }}^{2}}\theta }}$

$\displaystyle \therefore \ \begin{array}{|l|} \hline \displaystyle {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \\ \hline\end{array}$

$\displaystyle \ \ \ \text{Since}\ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$

$\displaystyle \ \ \ \frac{{{{{\sin }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}=\frac{1}{{{{{\sin }}^{2}}\theta }}$

$\displaystyle \therefore \ \begin{array}{|l|} \hline \displaystyle 1+{{\cot }^{2}}\theta ={{\operatorname{cosec}}^{2}}\theta \\ \hline\end{array}$